Solveeit Logo
Login

Question

Question: An ideal gas expands in volume from \(1\times {{10}^{-3}}\) \({{m}^{3}}\) to \(1\times {{10}^{-2}}\)...

An ideal gas expands in volume from 1×10−31\times {{10}^{-3}} m3{{m}^{3}} to 1×10−21\times {{10}^{-2}} m3{{m}^{3}} at 300K300K against a constant pressure of 1×1051\times {{10}^{5}} Nm−2N{{m}^{-2}} . The work done is:
A. −900J-900J
B. −1000J-1000J
C. 270kJ270kJ
D. −900kJ-900kJ

Explanation

Solution

ideal gas is defined as the gases in which collisions between the atoms are elastic and there are no intermolecular attractive forces. The ideal gas law is defined as the product of pressure and volume of a molecule is equal to the product of gas constant and temperature. Ideal gas is a hypothetical gas that tells us about the behaviour of gases under many conditions.
To calculate this question, we need to know about the relation between pressure, work and change in volume. The work calculated is equal to the product of negative external pressure and change in volume.

Formula used: W=−Pext(V2−V1)W=-{{P}_{ext}}\left( {{V}_{2}}-{{V}_{1}} \right)
Where, WW is the work done
Pext{{P}_{ext}} is the external pressure
V2{{V}_{2}} is the final volume
V1{{V}_{1}} is the initial volume

Complete step by step answer:
Here, it is given that the final volume is 1×10−2m31\times {{10}^{-2}}{{m}^{3}}
The initial volume is 1×10−3m31\times {{10}^{-3}}{{m}^{3}}
Temperature: 300K300K
Constant pressure: 1×105Nm−21\times {{10}^{5}}N{{m}^{-2}}
To calculate work done:
W=−Pext(V2−V1)W=-{{P}_{ext}}\left( {{V}_{2}}-{{V}_{1}} \right)
Where, WW is the work done
Pext{{P}_{ext}} is the external pressure
V2{{V}_{2}} is the final volume
V1{{V}_{1}} is the initial volume
Now, we substitute the value in the given formula we get,
W=−1×105(1×10−2−1×10−3)W=-1\times {{10}^{5}}\left( 1\times {{10}^{-2}}-1\times {{10}^{-3}} \right)
W=−1×105(10×10−3−1×10−3)W=-1\times {{10}^{5}}\left( 10\times {{10}^{-3}}-1\times {{10}^{-3}} \right)
On further solving we get,
W=−1×105(9×10−3)W=-1\times {{10}^{5}}\left( 9\times {{10}^{-3}} \right)
W=−900JW=-900J
Therefore, the work done for ideal gas expansion is −900J-900J

So, the correct answer is Option D .

Additional information: Ideal gas law is the hypothetical equation of ideal gas. It tells us about the behaviour of gases under many conditions. This law does not tell us about whether the gas heat or cools during expansion and compression.Ideal gases are the point masses that move in random straight line motion.Ideal gas law is the combination of simple gas law that is boyle’s law, Charles law and Avogadro's law. One pascal cubic metre is equal to the one joule.

Note: If work is done on the system, then work done is positive and if work is done by the system, then work done is negative. The work done in terms of pressure and volume is done by frictionless piston and ideal gases.